Partial Derivatives & Tangent Planes

Partial Derivatives

The partial derivative of a multivariate function ${f(x,y)}$ wrt ${x}$ is ${\frac{\partial f}{\partial x}\equiv f_x}$ . Similarly we can differentiate again to obtain the 2nd order partial derivative ${\frac{\partial^2 f}{\partial x^2}\equiv f_{xx}}$ . When we take the partial derivative wrt ${x}$ , we treat any other variables ${y,z}$ etc. as constants, for example if ${f(x,y)=x^2+y^2}$ then ${f_x = 2x}$ .

Formally, ${f: \mathbb{R}^2\rightarrow \mathbb{R}}$ the partial derivatives of ${f}$ at ${x_0, y_0}$ are:

$\displaystyle \left. \frac{\partial f}{\partial x} \right|_{x_0,y_0}=\lim_{h\rightarrow 0}\frac{f(x_0+h,y_0)-f(x_0,y_0)}{h}$

We can also take mixed partial derivatives, for example ${f_{xy}=\frac{\partial^2 f}{\partial y \partial x}=\frac{\partial}{\partial y} (\frac{\partial f}{\partial x})}$ .

Equality of mixed partial derivatives
– If all 2nd order mixed partial derivatives of ${f: \mathbb{R}^2\rightarrow \mathbb{R}}$ are continuous then ${f_{xy}=f_{yx}}$ .
– Continuous at a point ${(a,b)}$ if ${\lim_{(x,y)\rightarrow (a,b)} f(x,y)=f(a,b)}$ .

When we take a derivative of a function with one variable and evaluate it at a point it gives us the slope of the function at that point. Geometrically a partial derivative wrt ${x}$ is the slope of the surface in the ${x}$ direction at that point.

Tangent Planes

There are several ways to find a tangent plane to a surface. Firstly note that given a point ${(x_0,y_0,z_0=f(x_0,y_0))}$ on a surface, the two direction vectors ${(1,0,f_x(x_0,y_0))}$ and ${(0,1,f_x(x_0,y_0))}$ lie in the tangent plane.

Parametrically the tangent plane is ${\mathbf{r}=(x_0,y_0,z_0)+s(1,0,f_x(x_0,y_0))+t(0,1,f_x(x_0,y_0))}$ , where ${s,t}$ are scalars.

Alternatively take ${\mathbf{r_0}=(x_0,y_0,f(x_0,y_0))}$ , ${\mathbf{n}=(1,0,f_x(x_0,y_0))\times (0,1,f_x(x_0,y_0))=(f_x(x_0,y_0),f_y(x_0,y_0), -1)}$ (the cross product of two direction vectors in the plane gives a vector normal to the plane) and ${\mathbf{r}=(x,y,z)}$ . Point-normal form gives ${(\mathbf{r}-\mathbf{r_0})\cdot \mathbf{n} = 0}$ . We either substitute the vectors in directly to obtain:

$\displaystyle z=z_0+f_x(x_),y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$

Or note ${(\mathbf{r}-\mathbf{r_0})\cdot \mathbf{n} = 0 \implies \mathbf{r\cdot n}=\mathbf{r_0 \cdot n}}$ which gives:

$\displaystyle (x,y,z)\cdot((f_x(x_0,y_0),f_y(x_0,y_0), -1)=(x_0,y_0,f(x_0,y_0))\cdot((f_x(x_0,y_0),f_y(x_0,y_0)$

Find a tangent plane to ${f(x,y)=x^3-y^2}$ at (1,3)

The two direction vectors are ${(1,0,f_x(1,3))=(1,0,3)}$ and ${(0,1,f_y(1,3))=(0,1,-6)}$ . ${z_0=f(1,3)=-6}$ . Parametrically the tangent plane is:

$\displaystyle \mathbf{r}=(1,3,-6)+s(1,0,3)+t(0,1,-6)$

Alternatively the normal vector is ${(f_x(1,3),f_y(1,3), -1)=(3,-6,-1)}$ , so ${\mathbf{r\cdot n}=\mathbf{r_0 \cdot n}}$ gives:

$\displaystyle (x,y,z)\cdot((3,-6,-1)=(1,3,-6)\cdot(3,-6,-1)$

$\displaystyle 3x-6y-z=-9\ \text{ (scalar form)}$

The parametric equation and the scaler form are two equations for the same plane.

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